Integral Theorems of Vector Analysis

Abstract

The integral theorems of vector analysis build a relation between differentiation and integration and reduce often the ‘dimension of integration’ \(\bullet \) the area and geometrical center of planar figures (theorem of Green), \(\bullet \) the flux through a surface or volume (theorem of Gauß), \(\bullet \) or the circulation within a (curved) surface (theorem of Stokes).

Evaluate surface and volume integral of the theorem of Gauß for the flux of the vector field \(\varvec{F} = -x\,\hat{\varvec{i}}+ y\,\hat{\varvec{j}}+ 6z\,\hat{\varvec{k}}\) through the torus \(\mathcal {T}\). The surface of the torus is given by

For parametrization, consider the plane z\(=\)0 and express the non-centered circle in polar coordinates.

49.

Given a regular tetraeder \(\mathcal {T}\) with the corner points \(\{\varvec{A},\varvec{B},\varvec{C},\varvec{D}\}\), where

all triangles are equilateral with the sides \(\ell = \sqrt{3}\),

the triangle \(\Delta \varvec{ABC}\) lies in the plane \(z=0\),

the point \(\varvec{A}\) has the coordinates \(\begin{pmatrix} 1,&0,&0 \end{pmatrix}^\top \),

and the point \(\varvec{D}\) is on the positive z-axis.

(a)

Determine the coordinates of \(\{\varvec{B},\varvec{C},\varvec{D}\}\).

(b)

Move the whole tetraeder – without rotations – along the \(x-\)axis, so that in the shifted tetraeder \(\mathcal {T}_s\) the x-component of \(\varvec{C}_s\) is zero, while the y- and z-coordinates remain unchanged.

(c)

Determine the flux through the tetraeder \(\mathcal {T}_s\) for the vector field

Determine the curl and the divergence of the vector field \(\varvec{G}\) and use the result to figure out the circulation within the ellipse defined by the intersection of the plane \(\mathcal {E}\) and the cone \(\mathcal {C}\).

(b)

Calculate the flux through the smaller Dandelin sphere \(\mathcal {D}_1\), which is tangential to the cone and the plane.

The radius can be found in the plane y\(=\)0 by the incircle formula\(R= \frac{2A}{u}\)with the area A and the perimeter u of the corresponding triangle.

52.

Calculate the flux through the area \(\mathcal {O}_p = \left\{ \varvec{x}\in \mathbb {R}^3\!\!:\!z = \frac{1}{3} \sqrt{9-x^2-y^2}, |y|\le 1\right\} \) with the vector field

Given an arbitrary planar surface \(\mathcal {S}_\mathcal {E}\) in the plane \(\mathcal {E}: y+4x -2z = 0\). Find a non-conservative vector field \(\varvec{F}\) so that the circulation for all figures \(\mathcal {S}_\mathcal {E}\) is zero. The solution is not unique.

The planar figure\(\varvec{\Psi } = \begin{pmatrix} \cos ^3 t,&\sin ^3 t\end{pmatrix}^\top \)is known asOpen image in new window but also as cubocycloid, or paracycle.

In a geometrical definition, the astroid is created by tracing a marked point on a circle, which is rolling inside a fixed circle. Therefore, the curves belongs to the family ofOpen image in new window

The algebraic equation of the curve is\(x^{2/3}+y^{2/3} = 1\). Therefore, the curve belongs to the family ofOpen image in new window

Due to the symmetry, it is also possible to integrate only in the first quadrant (Fig.4.1).

The figure\(x^3+y^2-3xy = 0\) – in particular its closed part – is known asOpen image in new windowand the relevant part of the curve is visualized in Fig.4.2.

Inexercise 28, the x- and y-components are exchanged. Due to its symmetry, the result is the same curve (after the parametrization\(x = y\cdot t\)).

The parametrization by\(y =xt\) or \(x=yt\)might work for curves which contain the origin\(\varvec{0}\). In the area calculation by Green’s theorem, the calculation of\(\{x,\dot{x}\}\)or\(\{y,\dot{y}\}\)can be avoided in this case.

4.39. Area Enclosed by the Curve \(2(x^2-y^2) = (x^2+y^2)^2\)

Rational Parametrization

The point \(\varvec{0}\) fulfills the equation of the curve. Hence, we try again a parametrization by \(y = x\cdot t\):

This formula holds for closed curves which are either passing the origin or going around the origin. In particular it holds for ‘sectors’, which are defined by one curved line\(\rho (\phi )\)and two straight lines intersecting in the origin (cf. Fig.4.4).

4.40. Area Enclosed by \(\rho = \cos n \phi \)

is closed. For each choice of \(n\in \mathbb {N}\) the closed figure will have a blossom-like shape with different widths of the petals (cf. Fig. 4.5). We assume \(\phi _0=0\) as the starting point with the coordinates \(\varvec{\Psi }(0) = \begin{pmatrix}1,&0\end{pmatrix}^\top \). The curve will be closed when this point is reached again, which is equivalent to \(\phi = k\pi \) with \(k\in \mathbb {N}\).

4.42. Geometrical Center

When we remove the parts below the line \(x=y\), we obtain the interval \(t\in \left[ \frac{\pi }{4},\frac{5\pi }{4}\right] \) for the boundary part \(\partial B_1\). In addition we consider the straight line \(y = x\) in the interval \(y\in \left[ -2^{3/2}, 2^{3/2}\right] \) as boundary \(\partial B_2\):

The complete astroid is symmetric w.r.t. the axes \(x=0\), \(y=0\) and \(x = y\) and \(x = -y\). After removing the part below the line \(x = y\), the remaining figure is still symmetric w.r.t. to the line \(y = -x\). Therefore, we get the relation \(\bar{y} = - \bar{x}\) for the geometrical center (Fig. 4.7).

The first solution describes a circle with the radius \(\rho _1 = \sqrt{1+\sqrt{2}}\) and the enclosed area \(A =\rho _1^2 \pi = \left( 1+\sqrt{2}\right) \pi \).

The second solution might be an artifact as the squared radius \(\rho ^2_2 = 1-\sqrt{2}\) should not be negative. In previous examples, it was possible to interpret a negative radius as the reflected curve. We insert the radius \(\rho _2 = \sqrt{|1-\sqrt{2}|} = \sqrt{\sqrt{2}-1}\) into the equation and obtain

Due to symmetry, the curve is closing somewhere on the x-axis, which is equivalent to the condition \(t(t^2-3) = 0\), i.e. \(t = \pm \sqrt{3}\) or \(t = 0\). Now we consider, that the path of integration must be anti-clockwise enclosing the area. Hence, the integration interval starts at \(t=\sqrt{3}\):

A mathematical cylinderis defined by a planar curve\(\varvec{\Psi }\)and its translation\(\varvec{\Psi }_{\varvec{v}}\)in an arbitrary direction\(\varvec{v}\)outside the plane. The straight lines between\(\varvec{\Psi }\)and\(\varvec{\Psi }_{\varvec{v}}\)form the surface of the cylinder.The curve \(\varvec{\Psi }\)is not necessarily closed. If the curve is closed, then also the whole body is sometimes called a cylinder in the mathematical sense.

The divergence \(\mathop {\mathrm {div}}\nolimits \varvec{F} = 6\) is independent of the position, which makes the problem equivalent to determining the volume of the torus in Fig. 4.10. It is possible, to work with cylindrical coordinates, but we want to practice the volume integral with the Jacobian determinant here as well.

For a volume integral, a third parameter \(\xi \in [0,1]\) has to be introduced:

4.47. Flux Through a Semisphere in Spherical Coordinates

The normal vector\(\varvec{N}\)of a sphere around the origin is proportional to the radial vector\(\,\hat{\varvec{h}}_r\)with\(\varvec{N} = r^2\sin \vartheta \,\hat{\varvec{h}}_r\).

In case of surface integral over a (partial) spherical surface around the origin, the field components\(\,\hat{\varvec{h}}_\lambda \)and \(\,\hat{\varvec{h}}_{\vartheta }\)cannot contribute due to orthogonality of the frame vectors.

which is depending only on the z-coordinate. Hence, a volume integral within the upper Vivani’s figure (cf. Fig. 4.11) might be easier to evaluate. We express the shifted cylinder in polar coordinates. In the xy-plane we find

4.49. Flux Through a Tetraeder

a. Determine the Corner Coordinates

Based on the given conditions, we have to determine the corner points of the tetraeder. For \(\varvec{D}\) we use the known distance to \(\varvec{A}\) and obtain \(\varvec{D} = \begin{pmatrix} 0,&0,&\sqrt{2} \end{pmatrix}^\top \).

For \(\varvec{B},\varvec{C}\) we rotate the location \(\varvec{A}\) around the z-axis by the angle \(\phi =\pm \frac{2\pi }{3}\):

If we want to solve the problem by surface integrals, then we have to calculate the flux through four faces of the tetraeder. Therefore, we consider the parametrization of an arbitrary triangle \(\Delta (\varvec{P}_2,\varvec{P}_1,\varvec{P}_3)\) in space

4.50. Mathematical Cylinder

shows that the principle of Cavalieri could be applied again and we don’t need a complete parametrization of body visualized in Fig. 4.12. The area was determined already in exercise 37 with \(A(z) = \frac{3\pi }{8}\). The maximum height is given by the question and so we obtain the flux

The curl of the vector field is the null vector. Hence, the circulation within the ellipse – or any other surface – is also zero.

b. Flux Through Dandelin Sphere

This question has the remarkable property, that we can integrate the flux without knowing the exact parametrization like radius or center coordinates.

Cylinder Coordinates with Unknown Radius

Due to symmetry, the center of the sphere is somewhere on the z-axis with the coordinates \(\varvec{Z} = \begin{pmatrix} 0,&0,&z_0 \end{pmatrix}^\top \). We assume a radius of R, introduce the abbreviation \(\sigma = \sqrt{R^2-\rho ^2}\) and solve the problem in cylindrical coordinates:

The lattices have the dimensions \(b=|\varvec{C'A'}| = \frac{4}{5}\), \(a = |\varvec{C'B'}|= \frac{3}{5}\) and \(c = |\varvec{A'B'}| =1\). Hence we calculate the perimeter \(u = \frac{12}{5}\) and the area \(A =\frac{6}{25} \), which leads to the radius \(R = \frac{2A}{u} = \frac{1}{5}\). For the z-component of the center we obtain \(z_{0} = \sqrt{2} R = \frac{\sqrt{2}}{5}\).

After returning to the 3D space, we determine the center \(\varvec{Z} = \begin{pmatrix} 0,&0,&\frac{\sqrt{2}}{5} \end{pmatrix}^\top \) and the radius \(R = \frac{1}{5}\). Hence, we insert the geometry to the flux integral and obtain

As an alternative, we could solve the integration in spherical coordinates. We want to integrate a sphere centered around the origin \(\varvec{0}\) for simplicity. Therefore, we ‘translate’ the divergence of the vector field in the opposite direction

The oblate spheroidal coordinates represent also ellipsoid surfaces for all constant pairs \(\{p,\alpha \}\). Our calculation will be very simple, if we find a value of \(\alpha \) so that the surface \(\mathcal {O}_P\) is a coordinate surface.

Hence, we conclude that in the oblate spheroidal coordinate system, the parameter \(p = \sqrt{8}\) and the coordinate \(\alpha = {\mathop {\text {artanh}}}\, \frac{1}{3}\) represent the ellipsoidal surface with the equation

For every orthogonal coordinate system, the normal vector of a coordinate surface (\(q_i = \mathop {\mathrm {const.}}\nolimits \)) is parallel to the corresponding frame vector\(\,\hat{\varvec{h}}_{q_i}\). Assuming\(q_i = \beta \), the normal vector is found by the non-normalized frame vectors:

4.53. Flux Through a Paraboloid

fulfills the equation \(x^2+y^2-1 = \alpha ^2-1 = 2z\). This is equivalent to the coordinate surface \(\beta \equiv 1\) with \(u = \alpha \) and \(w = \gamma \) in parabolic coordinates. In other words, the paraboloid in Fig. 4.16 is a coordinate surface of the applied coordinate system.

We use the ‘frame vectors’ of exercise 14 and obtain the normal vector

This exercise demonstrates the benefit of the ‘optimal’ coordinate system. For known ‘frame vectors’ and coordinate surfaces, the calculation of the normal vector is performed in one line only.

4.54. ‘Theorem of Gauß’ for Surface Integrals

The surface \(\tilde{\mathcal {C}}\) is defined by a rotation of a one-dimensional function, which enables a parametrization in cylindrical coordinates with a variable radius \(\rho (z)\). Nevertheless, the integrals might not be handy.

To apply the theorem of Gauß, we calculate the divergence \(\mathop {\mathrm {div}}\nolimits \varvec{F} = 2-4+2 = 0\). Therefore, every closed volume will have a total flux of zero through its total surface.

The rotational surface \(\tilde{\mathcal {C}}\) is not closed, but open on both ends (Fig. 4.17). We add the missing partial areas and calculate the flux through them. For simplicity we choose circles in the plane \(z= 0\) and \(z=6\). The normal vector of these circles is \(\varvec{N} = \rho (\pm \,\hat{\varvec{k}})\) with the radius \(\rho \) depending on the height.

For the ‘bottom’ of the surface \(\tilde{\mathcal {C}}\) we find \(\rho = 3\) and the flux

We introduce the obvious parametrization \(\mathcal {S} = \begin{pmatrix} x,&4-z^2,&z \end{pmatrix}^\top \) with \(x\in [0,1]\) and \(z\in [-2,2]\) for the surface, which is visualized in Fig. 4.18. Then we calculate the normal vector

By interchanging the parameters – here x and y – the opposite direction of the normal vector is used. Without further condition (e.g. ‘pointing away from ...’), also the negative answers are correct in Stokes’ theorem.

b. Circulation via the Line Integral

For the line integral, we split the boundary into four parts.

We start with the boundary \(\varvec{\Psi }_1: \begin{pmatrix} t,&0,&2 \end{pmatrix}^\top \) with \(t\in [0,1]\) and obtain

This figure is a pattern within a square domain (Fig. 4.20, right). It is possible, but not recommended, to solve the question in cylindrical coordinates. The radius can be found by \(\rho = \frac{1}{|\cos \phi |+|\sin \phi |}\) according to exercise 3. We prefer to re-write into Cartesian coordinates. For the vector field, we obtain

We choose the surface integral with the normal vector \(\varvec{N} =\sin \vartheta \,\hat{\varvec{h}}_r\) of the unit sphere. Due to later multiplication with the normal vector, we have to derive only the \(\varvec{h}_r\)-component of the curl:

The vector field depends on the angles but not on the distance to the origin. The planar triangle is described by the same angles as the spherical one. Therefore, the circulation within the planar triangle must be the same \(\Omega _{\Delta \varvec{ABC}} = -3\).

4.59. Circulation within a Parabolic Partial Area

The domain \((x-2)^2+4(y+1)^2 = 1\) describes an ellipse in the plane \(z=0\) with the center \((2,-1)\). The semi-major axis \(a=1\) and the semi-minor axis \(b = \frac{1}{2}\) are oriented parallel to the coordinate axes (cf. Fig. 4.21). Its projection on the paraboloid \(z=x^2+y^2\) has the boundary

4.60. No Circulation for Planar Figures

Based on the Stokes’ theorem we conclude that the circulation will vanish when the product of normal vector and curl is zero for every figure. The normal vector of the plane \(y+4x -2z = 0\) is \( \varvec{N} = \begin{pmatrix} 4,&1,&-2 \end{pmatrix}^\top \). Hence, we have to determine a vector field whose curl is orthogonal to \(\varvec{N}\).

This problem is under-determined and we can use different methods to find an arbitrary solution. The methods and the results will differ in complexity. To keep the calculation simple, we assume a vector field with linear components

Due to the curl operator, every linear or non-linear function \(g_1(x)\) is eliminated for the first component, and in a similar way also functions of \(g_2(y)\) and \(g_3(z)\) for the second and third components, respectively. For the other variables we make an ansatz of a (linear) vector field

Although the vector field is non-conservative, we found two families of surfaces with vanishing circulation:

All partial areas on the surface \(\alpha = \mathop {\mathrm {const.}}\nolimits \) have zero circulation, because of the orthogonality between the surface’s normal vector and the curl of the vector field.

On the surface \(\beta = \mathop {\mathrm {const.}}\nolimits \), the circulation vanishes due to symmetry of \(\sin \gamma \) for the interval \(\gamma \in [0,\pi ]\).